logarithmic integral function and asymptotic expansion

Show that Logarithmic integral function $$\int_2^x {1\over \log(t)} \, dt = Li(x)$$ has asymptotic expansion of the form $${x\over \log(x)}\cdot\sum_{j=0}^\infty a_j\cdot (\log(x))^{-j}.$$

I tried different stuff but I did not conclude to solve it. Any help for solving this?

This is a very sketchy answer!

Observe, that

$$\text{Li(x)}=-\int_{\log(2)}^{\log(x)}\frac{e^{-y}}{y}=\text{Ei}(\log(x))-\text{Ei}(\log(2))$$

where $\text{Ei(x)}=\int_{x}^{\infty}\frac{e^{-q}}{q}$ the Expontential integral .

Doing intgration by parts N times we obtain $$ \text{Ei(x)}=\frac{e^{-x}}{x}-\frac{e^{-x}}{x^2}+\frac{2!e^{-x}}{x^3}+....+(-1)^{N-1}\frac{e^{-x}(N-1)!}{x^n}+R(N) $$ where $R(N)$ is the remaining integral.

We conclude that $$ \text{Ei(x)}=\frac{e^{-x}}{x}\sum_{k=1}^{\infty}(-1)^{k+1}\frac{k!}{x^k} $$

in an asymptotic sense. Now plug in $\log(x)$ and $\log(2)$ and you are done.